Re: Finite fields

What does this tell you about possible additional members to your new field?

-Dan

this makes no sense at all: 1/2 is undefined in a field of characteristic 2 (such as Z2).

instead, let's factor x4 - x into irreducible factors:

x4 - x = x(x - 1)(x2 + x + 1)

it is obvious that the linear factors x and x - 1 are irreducible (and have the roots x = 0 and x = 1 in Z2), to show that x2 + x + 1 is irreducible over Z2, it is sufficent to show it has no linear factors, that is: no roots in Z2:

02 + 0 + 1 = 0 + 1 = 1 ≠ 0
12 + 1 + 1 = (1 + 1) + 1 = 0 + 1 = 1 ≠ 0.

let u be a root of x2 + x + 1. just to be clear, what we are REALLY doing is taking the quotient ring:

we embed the field Z2 in this quotient by mapping a to the coset a + I (where we are taking I = (x2 + x + 1) because i'm lazy and hate typing a polynomial over and over again), for any element a in Z2 (there's only 2 of them)).

in this quotient ring u is the coset of x. it is clear that this actually gives a root:

I = 0 + I (the identity of the ring Z2[x]/I), which we are just going to call "0".
1 + I , which we are just going to call "1" (this is indeed the multiplicative identity of Z2[x]/I, verify this!).
x + I, which we are going to call "u"
(x+1) + I, which we are going to call "u+1".

note that for any polynomial of degree 2 or more, say f(x), we can turn it into one of the 4 cosets above by writing:

and the polynomials 0,1,x, and x+1 are the only elements of Z2[x] of degree < 2, together with 0.

now, we can use some "high-powered" ring theory, to establish that {0,1,u,u+1} is a field by showing that (x2+x+1) is a maximal ideal, because x2+x+1 is irreducible over Z2, but we can just show that our 4-element set is a field DIRECTLY.

it is immediate that Z2[x]/I is a ring, being a quotient ring. so we only need to verify that the set {1,u,u+1} forms a group under multiplication.

since we already know (since we're in a ring) the multiplication is associative, we only need to show closure, and inverses.